What is the solution for 2xy^2+x^2y^'=y^2 ?

The solution for 2xy^2+x^2y^'=y^2 is y=-x/(-1+2x-2xln(x)+c_{1)x}

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2xy^2+x^2y^'=y^2

2 x y^{2} + x^{2} y^{^{'}} = y^{2}

y = - \frac{x}{- 1 + 2 x - 2 x ln ( x ) + c _{1} x}

Solution steps

2 x y^{2} + x^{2} y^{'} (x) = y^{2}

Solve separable ODE:

y = - \frac{x}{- 1 + 2 x - 2 x ln ( x ) + c _{1} x}

y = - \frac{x}{- 1 + 2 x - 2 x ln ( x ) + c _{1} x}

y^{^{'}} = y (3 - y), y (0) = 5

f (x) = 3 sin (x) + cos (x)

x \to - \infty lim (x, - 2)

\int \frac{- 4}{( 3 x ) ^{3}} d x

y^{^{'}} = \frac{x ^{2} + y ^{2}}{x y}

What is the solution for 2xy^2+x^2y^'=y^2 ?
The solution for 2xy^2+x^2y^'=y^2 is y=-x/(-1+2x-2xln(x)+c_{1)x}